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\topic{Lecture 2.1/2.2 \\Matrix\\ \scriptsize Elementary Transformation, Rank (03,04 Nov 2009)}
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\section{Elementary Transformation}
The following are the elementary transformations  in a matrix
\begin{itemize}
\item Interchange of any two rows (or columns). If $ith$ row is interchanged with $jth$ row then this transformation is denoted by $R_{ij}$.
\item Multiplication of the elements of any rows (or columns) by a non-zero number. If each element of $jth$ row is multiplied by a non-zero scalar $k$, then this transformation is denoted by $k.R_j$.
\item Addition to the elements of any row(or column) the corresponding elements of any other row (or column) multiplied by a number. This transformation is denoted as $(R_i+kR_j)$, if each element of $jth$ row is multiplied by $k$ and then this quantity is added to $ith$ row.
\end{itemize}

If we transform a matrix $A$ to another matrix $B$ by means of elementary transformations, then the two matrices $A$ and $B$ are said to be equivalent. We denote $A \sim B$.

\subsection{Rank of a Matrix}
Consider a matrix $A$ of $m$ rows and $n$ columns, i.e. of order $m \times n$. We can find a square matrix by deleting some rows or some columns or both. Such obtained matrix, we call, Sub matrix of $A$. The determinant of this sub matrix is called Minor of the matrix. Now if we delete any number of rows or/and columns to get a non-vanishing minor, then the order of this minor, if highest, is called the Rank of the matrix $A$. Thus, in concise we may say that,

\begin{quote}
The rank of the matrix $A$ is said to be $r$ if, it has atleast one non-zero \textbf{minor} of order $r$ and every minor of $A$ of order higher than $r$ vanishes.
\end{quote}

\begin{quote}
If by using elementary transformation, we reduce the matrix in Echelon Form (Upper or Lower triangular form), then number of non zero rows is called rank of the that matrix.
\end{quote}
\begin{example}
Find the rank of the matrix
\[
A = \left[
\begin{array}[pos]{rrrr}
1 & 2 & 3 & 2  \\
2 & 3 & 5 & 1  \\
1 & 3 & 4 & 5  \\
\end{array}
\right]
\]
\end{example}
\paragraph{Solution:}  Apply $R_2 \rightarrow R_2 - 2 R_1$, $R_3 \rightarrow R_3-R_1$, we get
\[
A  \sim \left[
\begin{array}[pos]{rrrr}
1 & 2 & 3 & 2  \\
0 & -1 & -1 & -3  \\
0 & 1 & 1 & 3  \\
\end{array}
\right]
\]
Apply $R_3 \rightarrow R_3+R_2$, we get

\[
A  \sim \left[
\begin{array}[pos]{rrrr}
1 & 2 & 3 & 2  \\
0 & -1 & -1 & -3  \\
0 & 0 & 0 & 0  \\
\end{array}
\right]
\]
Which is Echlon Form, it implies that rank of the matrix is - number of non-zero rows : 2
\subsection{Normal Form } By performing elementary transformation, any non zero matrix $A$ can be reduced to one of the following normal forms, 
\[
I_r~~~~[I_r~O]~~~~\left[\begin{array}[pos]{c}I_r  \\ O  \\ \end{array}\right] ~~~~\left[\begin{array}[pos]{cc}
I_r & O  \\
O & O  \\
\end{array}
\right]
\]

Where $I_r$ denotes the identity matrix of order $r$. This number $r$, so obtained is called the rank of the matrix $A$.
\begin{example}
Reduce the matrix $A$ in one of the normal form and hence deduce the rank of matrix.
\[
A = \left[
\begin{array}[pos]{rrrr}
1 & -1 & 2 & -3  \\
4 & 1 & 0 & 2  \\
0 & 3 & 1 & 4  \\
0 & 1 & 0 & 2  \\
\end{array}
\right]
\]
\end{example}
\paragraph{Solution:}  Apply $R_2 - 4 R_1$, $R_{2,4}$

\[
A  \sim \left[
\begin{array}[pos]{rrrr}
1 & -1 & 2 & -3  \\
0 & 1 & 0 & 2  \\
0 & 3 & 1 & 4  \\
0 & 5 & -8 & 14  \\
\end{array}
\right]
\]

Apply $R_3 - 3 R_2$, $R_4 - 5 R_2$, 

\[
A  \sim \left[
\begin{array}[pos]{rrrr}
1 & -1 & 2 & -3  \\
0 & 1 & 0 & 2  \\
0 & 0 & 1 & -2  \\
0 & 0 & -8 & 4  \\
\end{array}
\right]
\]

Apply $R_4 + 8 R_3$,

\[
A  \sim \left[
\begin{array}[pos]{rrrr}
1 & -1 & 2 & -3  \\
0 & 1 & 0 & 2  \\
0 & 0 & 1 & -2  \\
0 & 0 & 0 & -12  \\
\end{array}
\right]
\]

Apply $C_2+C_1$, $C_3-2C_1$, $C_4+3C_1$,

\[
A  \sim \left[
\begin{array}[pos]{rrrr}
1 & 0 & 0 & 0  \\
0 & 1 & 0 & 2  \\
0 & 0 & 1 & -2  \\
0 & 0 & 0 & -12  \\
\end{array}
\right]
\]

Apply $C_4-2C_2$, $C_4+2C_3$, $\frac{-1}{12}C_4$,

\[
A \sim \left[
\begin{array}[pos]{rrrr}
1 & 0 & 0 & 0  \\
0 & 1 & 0 & 0  \\
0 & 0 & 1 & 0  \\
0 & 0 & 0 & 1  \\
\end{array}
\right]~~=~I_4
\]
Hence, $r=4$, rank of matrix is $4$.
\subsection{Inverse of a Matrix}
Let $A$ be a non-singular square matrix of order $n$. Let $B$ be another sqare matrix of the same order such that
\[AB = BA =I\]
where $I$ is Identity matrix of order $n$. In such case $B$ is called inverse of $A$ and is denoted as $A^{-1}$, so that
\[AA^{-1} = A^{-1}A = I\]
\paragraph{Theorem}
If inverse of $A$ exists, then it is unique
Let $B$ and $C$ be the inverse of $A$, then by deefinition, we have 
\[AB = BA =I ~~~~~AC=CA=I\]
Now premultiply First term by C and use second relation
\[CAB = CI  \Rightarrow IB = CI \Rightarrow B = C\]
which implies that inverse is unique.

\begin{example}
Find the inverse of the matrix by using elementary transformation
\[
A = \left[
\begin{array}[pos]{rr}
1 & -1 \\
4 & 1 \\
\end{array}
\right]
\]
\end{example}
\paragraph{Solution:}
We may write $A = I A$,
\[
\left[
\begin{array}[pos]{rr}
1 & -1 \\
4 & 1 \\
\end{array}
\right] = \left[
\begin{array}[pos]{rr}
1 & 0 \\
0 & 1 \\
\end{array}
\right]
A
\]
Now use only row transformation, reduce the matrix on left to identity matrix. similar operation will convert Identity matrix on right into inverse of $A$  (REM: No column tranformation.)
Apply $R_2 \rightarrow R_2 - 4 R_1$
\[
\left[
\begin{array}[pos]{rr}
1 & -1 \\
0 & 5 \\
\end{array}
\right] = \left[
\begin{array}[pos]{rr}
1 & 0 \\
0 & 1 \\
\end{array}
\right]
A
\]
Apply $R_2 \rightarrow \frac{R_2}{5} $
\[
\left[
\begin{array}[pos]{rr}
1 & -1 \\
0 & 1 \\
\end{array}
\right] = \left[
\begin{array}[pos]{rr}
1 & 0 \\
\frac{-4}{5}& \frac{1}{5} \\
\end{array}
\right]
A
\]
Apply $R_1 \rightarrow R_1+R_2 $
\[
\left[
\begin{array}[pos]{rr}
1 & 0 \\
0 & 1 \\
\end{array}
\right] = \left[
\begin{array}[pos]{rr}
\frac{1}{5} & \frac{1}{5} \\
\frac{-4}{5}& \frac{1}{5} \\
\end{array}
\right]
A
\]
Thus inverse of $\left[
\begin{array}[pos]{rr}
1 & -1 \\
0 & 5 \\
\end{array}
\right]$
is $\frac{1}{5} \left[
\begin{array}[pos]{rr}
 1& 1 \\
-4& 1 \\
\end{array}
\right]$
\section*{Problems}
\begin{enumerate}
\item  \textbf{ }Reduce the following matrix to upper triangular form (Echelon form): 
\[\left[\begin{array}{rrr} {1} & {2} & {3} \\ {2} & {5} & {7} \\ {3} & {1} & {2} \end{array}\right]\]

\item   Transform $\left[\begin{array}{rrr} {1} & {3} & {3} \\ {2} & {4} & {10} \\ {3} & {8} & {4} \end{array}\right]$ into a unit matrix.

\item   Find the inverse of the following matrix employing elementary transformation: 
\[\left[\begin{array}{rrr} {3} & {-3} & {4} \\ {2} & {-3} & {4} \\ {0} & {-1} & {1} \end{array}\right]\]

\item   Find by elementary row transformation the inverse of the matrix$\left[\begin{array}{rrr} {0} & {1} & {2} \\ {1} & {2} & {3} \\ {3} & {1} & {1} \end{array}\right]$.  

\item   Compute the inverse of the following matrix by using elementary transformations
\[\left[\begin{array}{cccc} {2} & {-6} & {-2} & {-3} \\ {5} & {-13} & {-4} & {-7} \\ {-1} & {4} & {1} & {2} \\ {0} & {1} & {0} & {1} \end{array}\right]\]

\item   Find the inverse of the matrix.$\left[\begin{array}{cccc} {2} & {4} & {3} & {2} \\ {3} & {6} & {5} & {2} \\ {2} & {5} & {2} & {-3} \\ {4} & {5} & {14} & {14} \end{array}\right]$  

\item   Reduce the matrix A=$\left[\begin{array}{rrr} {-1} & {2} & {-2} \\ {1} & {2} & {1} \\ {-1} & {-1} & {0} \end{array}\right]$ to diagonal form. 

\item   Find the rank of $\left[\begin{array}{cccc} {1} & {2} & {-1} & {3} \\ {4} & {1} & {2} & {1} \\ {3} & {-1} & {1} & {2} \\ {1} & {2} & {0} & {1} \end{array}\right]$ by reducing it to normal form. 

\item   Reduce the matrix A to its normal form, when A 
\[=\left[\begin{array}{cccc} {1} & {2} & {-1} & {4} \\ {2} & {4} & {3} & {4} \\ {1} & {2} & {3} & {4} \\ {-1} & {-2} & {6} & {-7} \end{array}\right]\] 
Hence find the rank of A.  

\item    Find non-singular matrices P, Q so that PAQ is a normal form where 
\[A =\left[\begin{array}{rrr} {2} & {1} & {-3} \\ {3} & {-3} & {1} \\ {1} & {1} & {1} \end{array}\begin{array}{c} {-6} \\ {2} \\ {2} \end{array}\right]\]

\item    If A =$\left[\begin{array}{rrr} {3} & {-3} & {4} \\ {2} & {-3} & {4} \\ {0} & {-1} & {1} \end{array}\right]$ , find two non singular matrices $P$ and $Q$ such that $PAQ = I$. Hence find $A^{-1}$. 

\item   Find the rank of the matrix
\[ (i)  \left[\begin{array}{rrr} {1} & {2} & {\begin{array}{cc} {3} & {2} \end{array}} \\ {2} & {3} & {\begin{array}{cc} {5} & {1} \end{array}} \\ {1} & {3} & {\begin{array}{cc} {4} & {5} \end{array}} \end{array}\right]
~~~~~~~~~~~~(ii)	 \left[\begin{array}{cccc} {-1} & {2} & {3} & {-2} \\ {2} & {-5} & {1} & {2} \\ {3} & {-8} & {5} & {2} \\ {5} & {-12} & {-1} & {6} \end{array}\right] \]
\[
(iii) \left[\begin{array}{cccc} {3} & {-4} & {-1} & {2} \\ {1} & {7} & {3} & {1} \\ {5} & {-2} & {5} & {4} \\ {9} & {-3} & {7} & {7} \end{array}\right] ~~~~~~~~~~(iv) \left[\begin{array}{cccc} {2} & {3} & {-1} & {-1} \\ {1} & {-1} & {-2} & {-4} \\ {3} & {1} & {3} & {-2} \\ {6} & {3} & {0} & {-7} \end{array}\right]\]

\item   Use elementary transformation to reduce the following matrix A to triangular form and hence find the rank of A=$\left[\begin{array}{cccc} {2} & {3} & {-1} & {-1} \\ {1} & {-1} & {-2} & {-4} \\ {3} & {1} & {3} & {-2} \\ {6} & {3} & {0} & {-7} \end{array}\right]$. 


\item   Obtain a matrix N in the normal form equivalent to 
\[(i)~~\left[\begin{array}{rrr} {1} & {1} & {2} \\ {1} & {2} & {3} \\ {0} & {-1} & {-1} \end{array}\right] ~~~~(ii) \left[\begin{array}{cccc} {0} & {0} & {0} & {\begin{array}{cc} {0} & {0} \end{array}} \\ {0} & {4} & {5} & {\begin{array}{cc} {0} & {0} \end{array}} \\ {0} & {9} & {1} & {\begin{array}{cc} {-1} & {2} \end{array}} \\ {0} & {10} & {0} & {\begin{array}{cc} {1} & {11} \end{array}} \end{array}\right] \]
Hence find non singular matrices P and Q such that PAQ = N

\item   Using elementary transformations, reduce the following matrices to the canonical form (or row-reduced echelon form): 
\[ ( i) A = \left[\begin{array}{cccc} {0} & {0} & {0} & {\begin{array}{cc} {0} & {0} \end{array}} \\ {0} & {1} & {2} & {\begin{array}{cc} {3} & {4} \end{array}} \\ {0} & {2} & {3} & {\begin{array}{cc} {4} & {1} \end{array}} \\ {0} & {3} & {4} & {\begin{array}{cc} {1} & {2} \end{array}} \end{array}\right]
~~~~ (ii)  A = \left[\begin{array}{cccc} {0} & {4} & {-12\begin{array}{cc} {} & {8} \end{array}} & {9} \\ {0} & {2} & {-6\begin{array}{cc} {} & {2} \end{array}} & {5} \\ {0} & {1} & {-3\begin{array}{cc} {} & {6} \end{array}} & {4} \\ {0} & {-8} & {24\begin{array}{cc} {} & {3} \end{array}} & {1} \end{array}\right]\]

\item   Reduce the matrix 
$A = \left[\begin{array}{rrr} {1} & {2} & {1} \\ {-1} & {0} & {2} \\ {2} & {1} & {3} \end{array}\right]$ to $I_3$  by elementary transformations.

\item   Reduce the matrix  $\left[\begin{array}{rrr} {-1} & {2} & {\begin{array}{cc} {1} & {8} \end{array}} \\ {2} & {1} & {\begin{array}{cc} {-1} & {0} \end{array}} \\ {3} & {2} & {\begin{array}{cc} {1} & {7} \end{array}} \end{array}\right]$ to triangular form and show that the reduction of a matrix to triangular form is not unique.

\item   Transform $\left[\begin{array}{rrr} {1} & {1} & {3} \\ {1} & {3} & {-3} \\ {-2} & {-4} & {4} \end{array}\right]$ into a unit matrix.

\item   Using the Gauss-Jordan reduction method, find the inverse of the matrix 
\[\left[\begin{array}{rrr} {1} & {1} & {3} \\ {1} & {3} & {-3} \\ {-2} & {-4} & {-4} \end{array}\right]\]

\end{enumerate}

\end{document}




